TARTARIC ACID
35
found the monohydrate is calculated, 18 parts of water corresponding with
98 of H2S04. Finally, subtraction of the percentage of H2S04 from 100
gives the free S03.
(b) In -presence of SOZ: in this case the total acidity must be diminished
by that due to the sulphurous acid, that is, before calculating the S03 from
the acidity, the number of c.c. of N/io-iodine used in determining the
sulphurous acid is divided by 10 and the result subtracted from the number
of c.c. of N/2-soda used in the measurement of the. acidity x; from the
number thus obtained the S03 is calculated as in (a). The sum of the S03
thus found and of the S02 found directly (i c.c. N/io-iodine = 0-0032 gram
of SO 2) is subtracted from 100, the result being the water, from which the
H2SO4 is calculated as in the previous case. Lastly, the S03 is calculated
by subtracting from 100 the sum of the H2S04 and the S02.
Example (of case &) : 3-422 grams of fuming acid, dissolved in water, were
made up to 500 c.c. For 100 c.c. of this solution (= 0-6844 gram of substance)
30-10 c.c. of N/a-caustic soda were required, or 5-25 c.c. of N/io-iodine. Then
30-10 0-525 = 29-575,
so that the SO3 will be
29'575 X 0-02003 = 0-5924,
or, allowing for the amount of the fuming acid in 100 c.c.,
SO 3 = 86-56%.
On the other hand the sulphurous acid will be
5-25 X 0-0032 = 0-0168
i.e., ' S02= 2-46%.
Hence the water will be
loo (86-56 + 2-46) = 10-98.
To 10-98 of water there correspond 59-78 of H2SO4, since
18 : 98 : : 10-98 : 59-78.
The free SO3 will therefore be
100 (59-78 + 2'46) = 37'76-
The acid thus contains
H2S04 5978%
S03 37-76%
S02 2-46%
TARTARIC ACID
C4H6Oa = 150
Large, colourless, transparent, odourless, non-hygroscopic crystals of
acid taste, readily soluble in water or alcohol. It may be contaminated
especially by small quantities of sulphuric acid or sulphates, salts of calcium,
potassium, iron, copper, and particularly lead, and sometimes arsenic. It
may be adulterated with cream of tartar, potassium and sodium sulphates,
alum and oxalic acid.
The tests to be made are as follows :
1. Solubility.It should dissolve completely in water (absence of
calcium tartrate or sulphate).
1 The reason of this is that, when the acidity is determined in presence of sulphurous
acid with methyl orange as indicator, neutrality of the liquid is reached when the acid
sulphite, NaHSO3, and not the normal sulphite, Na2SO3, is formed. Consequently, i c.c.
of N/io-iodine solution is equivalent; not to o-t c.c., but to only 0-05 c.c. of N-soda and
therefore to o-i c.c. N/2-soda.is calculated as percentage, of